Question

Match the inverse functions in column I with their principal values in column II and choose the correct option from the codes given below.

Column I		Column II
A	${\cot }^{-1}\sqrt{3}$	1	$\frac{2\pi }{3}$
B	${\cos }^{-1}\left(-\frac{1}{\sqrt{2}}\right)$	2	$-\frac{\pi }{4}$
C	${\mathrm{cosec}}^{-1}(-\sqrt{2})$	3	$\frac{3\pi }{4}$
D	${\cos }^{-1}\left(\frac{1}{2}\right)+2{\sin }^{-1}\left(\frac{1}{2}\right)$	4	$\frac{\pi }{6}$

• A. A - (4), B - (3),C - (2), D - (1)
• B. A - (3), B - (4),C - (2), D - (1)
• C. A - (2), B - (3),C - (4), D - (1)
• D. A - (4), B - (3),C - (1), D - (2)

Answer 1

Answer: (A)

A. Let ${\cot }^{-1}(\sqrt{3})=\theta \Rightarrow \cot \theta =\sqrt{3}$
We know that, the range of principal value branch of ${\cot }^{-1}\theta$ is $(0,\pi )$.
$\because \cot \theta =\sqrt{3}=\cot \frac{\pi }{6}\Rightarrow \theta =\frac{\pi }{6}$ where, $\theta \in (0,\pi )$
$\Rightarrow {\cot }^{-1}(\sqrt{3})=\frac{\pi }{6}$
Hence, the principal value of ${\cot }^{-1}(\sqrt{3})$ is $\frac{\pi }{6}$.
B. Let ${\cos }^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\theta \Rightarrow \cos \theta =-\frac{1}{\sqrt{2}}$
We know that the range of principal value branch of ${\cos }^{-1}\theta$ is $[0,\pi ]$
$\because \cos \theta =-\frac{1}{\sqrt{2}}=-\cos \frac{\pi }{4}=\cos \left(\pi -\frac{\pi }{4}\right)$
$\Rightarrow [\because \cos (\pi -\theta )=-\cos \theta )]$
$\theta =\frac{3\pi }{4}\text{ where, }\theta \in [0,\pi ]$
$\Rightarrow {\cos }^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{3\pi }{4}$
Hence, the principal value of ${\cos }^{-1}\left(-\frac{1}{\sqrt{2}}\right)$ is $\frac{3\pi }{4}$.
C. Let ${\mathrm{cosec}}^{-1}(-\sqrt{2})=\theta \Rightarrow \mathrm{cosec}\theta =-\sqrt{2}$
We know that, the range of principal value branch of ${\mathrm{cosec}}^{-1}\theta$ is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]-\{0\}$.
$\because \mathrm{cosec}\theta =-\sqrt{2}=-\mathrm{cosec}\frac{\pi }{4}=\mathrm{cosec}\left(-\frac{\pi }{4}\right)$
$[\because \mathrm{cosec}(-\theta )=-\mathrm{cosec}\theta ]$
$\Rightarrow \theta =-\frac{\pi }{4}\text{ where, }\theta \in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]-\{0\}$
$\Rightarrow {\mathrm{cosec}}^{-1}(-\sqrt{2})=-\frac{\pi }{4}$
Hence, the principal value of ${\mathrm{cosec}}^{-1}(-\sqrt{2})$ is $-\frac{\pi }{4}$.
D. Let ${\cos }^{-1}\left(\frac{1}{2}\right)=x\Rightarrow \cos x=\frac{1}{2}=\cos \frac{\pi }{3}$
$\Rightarrow x=\frac{\pi }{3}\in [0,\pi ]$ (principal value branch)
Again, let ${\sin }^{-1}\left(\frac{1}{2}\right)=y$
$\Rightarrow \sin y=\frac{1}{2}=\sin \frac{\pi }{6}$
$\Rightarrow y=\frac{\pi }{6}\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ (principal value branch)
$\therefore {\cos }^{-1}\left(\frac{1}{2}\right)+2{\sin }^{-1}\left(\frac{1}{2}\right)=x+2y$
$=\frac{\pi }{3}+2\times \frac{\pi }{6}=\frac{2\pi }{3}$